91Ó°ÊÓ

The Newton’s law of gravitation is given by,

\({F_N} = G\frac{{mM}}{{{r^2}}}\)...... (i)

Here\(m\)and\(M\)are the masses,\(G\)is universal gravitational constant,\(r\)is the distance between them and\(F\)is the force.

The centripetal force is given by,

\({F_G} = m\frac{{{v^2}}}{r}\)....... (ii)

Here\(v\)is the speed.

On equating (i) and (ii) we get,

\(\begin{align}{F_N} &= {F_G}\\G\frac{{mM}}{{{r^2}}} &= m\frac{{{v^2}}}{r}\\M &= \frac{{{v^2}r}}{G}\end{align}\)

Therefore the expression for the mass of the galaxy is,

\(M = \frac{{{v^2}r}}{G}\)

Substitute\(6.67 \times {10^{ - 11}}\,{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}}\)for\(G\),\(2.7 \times {10^5}\,{\rm{m/s}}\)for\(v\)and\(30,000\,{\rm{ly}}\)for\(r\)in the above equation,

\(\begin{align}M &= \frac{{{{\left( {2.7 \times {{10}^5}\,{\rm{m/s}}} \right)}^2}\left( {30,000\,{\rm{ly}}} \right)\left( {9.46 \times {{10}^{15}}\,{\rm{m/ly}}} \right)}}{{6.67 \times {{10}^{ - 11}}\,{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}}}}\\ &= 3 \times {10^{41}}\,{\rm{kg}}\end{align}\)

Therefore the mass of the star in multiple of solar mass is,

\(\begin{align}M &= \frac{{3 \times {{10}^{41}}\,{\rm{kg}}}}{{1.99 \times {{10}^{30}}\,{\rm{kg}}}}\\ &= 1.5 \times {10^{11}}\end{align}\)

Therefore the mass of the star is \(3 \times {10^{41}}\,{\rm{kg}}\) and the mass of the star in multiples of solar mass based on the velocity of the star and distance is from center of the galaxy is \(1.5 \times {10^{11}}\).